y^{''}-3y^'+2y=4e^{et},y(0)=-3,y^'(0)=5

解

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{e t}, y (0) = - 3, y^{^{'}} (0) = 5

y = (- 3 + \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} - \frac{4}{e ^{2} - 3 e + 2}) e^{2 t} - \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} e^{t} + \frac{4}{e ^{2} - 3 e + 2} e^{e t}

解答ステップ

y^{''} (t) - 3 y^{'} (t) + 2 y = 4 e^{e t}

線形 ODE を解く:

y = (- 3 + \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} - \frac{4}{e ^{2} - 3 e + 2}) e^{2 t} - \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} e^{t} + \frac{4}{e ^{2} - 3 e + 2} e^{e t}

y = (- 3 + \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} - \frac{4}{e ^{2} - 3 e + 2}) e^{2 t} - \frac{11 e ^{2} - 37 e + 30}{e ^{2} - 3 e + 2} e^{t} + \frac{4}{e ^{2} - 3 e + 2} e^{e t}